(2.9) | ||

(2.10) |

It follows immediately from the normalization condition, Equation (2.5), that

so . The best known example of a two-state system is a tossed coin. The two outcomes are ``heads'' and ``tails,'' each with equal probabilities . So, for this system.

Suppose that we make statistically independent observations of . Let us determine the probability of occurrences of the outcome , and occurrences of the outcome 2, with no regard as to the order of these occurrences. Denote this probability . This type of calculation crops up very often in probability theory. For instance, we might want to know the probability of getting nine ``heads'' and only one ``tails'' in an experiment where a coin is tossed ten times, or where ten coins are tossed simultaneously.

Consider a simple case in which there are only three observations. Let us try to evaluate the probability of two occurrences of the outcome 1, and one occurrence of the outcome 2. There are three different ways of getting this result. We could get the outcome 1 on the first two observations, and the outcome 2 on the third. Or, we could get the outcome 2 on the first observation, and the outcome 1 on the latter two observations. Or, we could get the outcome 1 on the first and last observations, and the outcome 2 on the middle observation. Writing this symbolically, we have

(2.12) |

Here, the symbolic operator stands for ``and,'' whereas the symbolic operator stands for ``or.'' This symbolic representation is helpful because of the two basic rules for combining probabilities that we derived earlier in Equations (2.4) and (2.8):

(2.13) | ||

(2.14) |

The straightforward application of these rules gives

(2.15) |

for the case under consideration.

The probability of obtaining occurrences of the outcome in observations is given by

where is the number of ways of arranging two distinct sets of and indistinguishable objects. Hopefully, this is, at least, plausible from the previous example. There, the probability of getting two occurrences of the outcome 1, and one occurrence of the outcome 2, was obtained by writing out all of the possible arrangements of two s (the probability of outcome 1) and one (the probability of outcome 2), and then adding them all together.